How Much is the Change in Volume? (multivariable differentiation)

[BlBl]

The books says:

The gas equation for one mole of oxygen relates its pressure, P (in atmospheres), its temperature, T (in K), and its volume, V (in cubic decimeters, dm³):

\(\displaystyle T = 16.574 \dfrac{1}{V} - 0.52754 \dfrac{1}{V^2} - 0.3879P + 12.187V P\).

(a) Find the temperature T and differential d_T if the volume is 25 dm³ and the pressure is 1 atmosphere.
(b) Use your answer to part (a) to estimate how much
the volume would have to change if the pressure increased by 0.1 atmosphere and the temperature remained constant.

Now, I've answered (a), but I can't seem to make headway on (b). I've tried (and failed at) rearranging it to isolate V to one side of the equation. I've tried doing a tangent plane approximation to T(1,25)=304.95 but I haven't gotten the answer in the back of the book. I tried adjusting d_P to the new pressure of 1.1 atm to see if that would help, but that was a no go.

I eventually just broke out the ideal gas formula of pV=nRT and got a close answer to the one in the back of the book (I got change in V of ≈2.2597).

Help?

EDIT: Sorry, I accidentally had a small v rather than a capital V in one expression. That's fixed.

 

[Deleted member 4993]

The books says:



Now, I've answered (a), but I can't seem to make headway on (b). I've tried (and failed at) rearranging it to isolate V to one side of the equation. I've tried doing a tangent plane approximation to T(1,25)=304.95 but I haven't gotten the answer in the back of the book. I tried adjusting dP to the new pressure of 1.1 atm to see if that would help, but that was a no go.

I eventually just broke out the ideal gas formula of pV=nRT and got a close answer to the one in the back of the book (I got change in V of ≈2.2597).

Help?

What is the difference between

'V' (as in 16.574/V)

and

'v' (as in 0.52754/v²)?

In addition, please show your work on part (a).

 

[BlBl]

What is the difference between

'V' (as in 16.574/V)

and

'v' (as in 0.52754/v²)?

In addition, please show your work on part (a).

There was no difference. I accidentally had a small v rather than a capital V in one expression. That's fixed.

Here's the work on (a):

\(\displaystyle T = 16.574 \dfrac{1}{V} - 0.52754 \dfrac{1}{V^2} - 0.3879P + 12.187VP\)

\(\displaystyle V=25 & P=1\)

\(\displaystyle T = 16.574 \dfrac{1}{25} - 0.52754 \dfrac{1}{25^2} - 0.3879 + 12.187(25)\)

\(\displaystyle T = .66296 - 0.000844064 - 0.3879 + 304.675\)

\(\displaystyle T = 304.95\)

For the second part of (a)

\(\displaystyle d_{T} = T_{P}(P,V)d_{P}+T_{V}(P,V)d_{V}\)

\(\displaystyle d_{T}=(-0.3879+12.187V)d_{P}-(\dfrac{16.574}{V^2}+\dfrac{2(0.52754)}{V^3}+12.187P)d_{V}\)

\(\displaystyle d_{T}=(-0.3879+12.187(25))d_{P}-(\dfrac{16.574}{25^2}+\dfrac{2(0.52754)}{25^3}+12.187)d_{V}\)

\(\displaystyle d_{T}=(304.2871)d_{P}-(12.21358593)d_{V}\)

 

[Deleted member 4993]

There was no difference. I accidentally had a small v rather than a capital V in one expression. That's fixed.

Here's the work on (a):

\(\displaystyle T = 16.574 \dfrac{1}{V} - 0.52754 \dfrac{1}{V^2} - 0.3879P + 12.187VP\)

\(\displaystyle V=25 & P=1\)

\(\displaystyle T = 16.574 \dfrac{1}{25} - 0.52754 \dfrac{1}{25^2} - 0.3879 + 12.187(25)\)

\(\displaystyle T = .66296 - 0.000844064 - 0.3879 + 304.675\)

\(\displaystyle T = 304.95\)

For the second part of (a)

\(\displaystyle d_{T} = T_{P}(P,V)d_{P}+T_{V}(P,V)d_{V}\)

\(\displaystyle d_{T}=(-0.3879+12.187V)d_{P}-(\dfrac{16.574}{V^2}+\dfrac{2(0.52754)}{V^3}+12.187P)d_{V}\)

dT = 0 ← constant temperature

find dV when dP = 0.1 when V = 25 and P = 1

\(\displaystyle d_{T}=(-0.3879+12.187(25))d_{P}-(\dfrac{16.574}{25^2}+\dfrac{2(0.52754)}{25^3}+12.187)d_{V}\)

\(\displaystyle d_{T}=(304.2871)d_{P}-(12.21358593)d_{V}\)

.

 

[Bob Brown MSEE]

Part B: Numerical Solution

Does your book give 2.27477 for part B?

By the way, you have done an excellent job of showing your work. However it is equally important to share EVERYTHING that you have. The value in the book is extremely important to help us explain why there might be a difference. Thankyou!

Check: T[V,P] = T[25, 1] = T[25 - 2.27477, 1.1] = 304.949

 

[Bob Brown MSEE]

Part B

I think you have a sign error in Tv,

but then I get
0 = (304.2871)dp + (12.1605)dv
dv=-2.5

 

[BlBl]

.

Thanks Subhotosh ;-)

Someone at school showed me how to do it as well. So I'm good on this problem now.

It seems that none of the examples from the book dealt with a change in the total differential/derivative so when it came up I was at a loss as to how to approach it.

Now I'm less afraid of future problems with variations in d_f, d_x, or d_y. I'll still ask if I need anything.

 

[BlBl]

Does your book give 2.27477 for part B?

By the way, you have done an excellent job of showing your work. However it is equally important to share EVERYTHING that you have. The value in the book is extremely important to help us explain why there might be a difference. Thankyou!

Check: T[V,P] = T[25, 1] = T[25 - 2.27477, 1.1] = 304.949

Hey Bob,

Thanks for the compliments. I really try to work at presentation of my problems when I'm writing things here. Other people are taking the time to help me out so I might as well take the time to make it look good.

The value for (b) given in the book is "≈2.5". I just kept more precise values even if no such significant figures could be justified (though the book doesn't care about SF). The book also rounded the 304.949 to 304.9, the 304.2871 to 304.29, the 12.187 to 12.16. In fact that last value should be a tip that I did some thing wrong here as you pointed out in the above message about the sign error.

I think I distributed a negative where I should've just left things as they were. The answer should be positive as the volume will become greater to accommodate the pressure increase and maintain constant temperature.